1 + - ^2 ( 1 + ) × ^2 + cose c^2 + , 0° < θ < 90°, is equal to:

Answer: Option B

$$\begin{array}{rl}& \frac{1+\cos \theta -{\sin }^2\theta }{\sin \theta \left(1+\cos \theta \right)}\times \frac{\sqrt{{\sec }^2\theta +\text{cose}{\text{c}}^2\theta }}{\tan \theta +\cot \theta }\\ & \Rightarrow \frac{{\cos }^2\theta -\cos \theta }{\sin \theta \left(1+\cos \theta \right)}\times \frac{\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}}{\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }}\\ & \Rightarrow \frac{\cos \theta \left(\cos \theta +1\right)}{\sin \theta \left(1+\cos \theta \right)}\times \frac{\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }}}{\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }}\\ & \Rightarrow \frac{\cos \theta }{\sin \theta }\times \frac{\sqrt{\frac{1}{{\sin }^2\theta {\cos }^2\theta }}}{\frac{1}{\cos \theta \sin \theta }}\\ & \Rightarrow \cot \theta \times \frac{\frac{1}{\cos \theta \sin \theta }}{\frac{1}{\cos \theta \sin \theta }}\\ & \Rightarrow \cot \theta \end{array}$$